# Prove the following

Question:

If vectors $\overrightarrow{a_{1}}=x \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{a_{2}}=\hat{i}+y \hat{j}+z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i}+y \hat{j}+z \hat{k}$ is :

1. $\frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$

2. $\frac{1}{\sqrt{2}}(i-j)$

3. $\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$

4. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})$

Correct Option: , 3

Solution:

$\frac{x}{1}=-\frac{1}{y}=\frac{1}{z}=\lambda($ let $)$

Unit vector parallel to $x \hat{i}+y \hat{j}+z \hat{k}=\pm \frac{\left(\lambda \hat{i}-\frac{1}{\lambda} \hat{j}+\frac{1}{\lambda} \hat{k}\right)}{\sqrt{\lambda^{2}+\frac{2}{\lambda^{2}}}}$

For $\lambda=1$, it is $\pm \frac{(i-3+k)}{\sqrt{3}}$